Interesting question. I'd be interested to learn what led you to this question.
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Jim ConantFeb 25 '11 at 15:06

You have a misprint in your polynomial, right? It must be $P_{g}(x_{1}, \cdots,x_{n})$
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Pedro Martins RodriguesFeb 25 '11 at 15:15

The road to this question is not that long actually: all symmetric and translation-invariant polynomials are linear combinations of (a subset of) polynomials obtained from multigraphs. For example discriminants. As it turns out that about half of the graphs yield the zero polynomial, it was a natural question...
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Per AlexanderssonFeb 25 '11 at 16:55

1 Answer
1

I am afraid this innocuous-looking question is in fact extremely hard, and I would be surprised if one could find a necessary and sufficient criterion which is more useful than the definition itself. Here is why:

The question pertains to the classical invariant theory of binary forms.
Firstly, suppose your graph is $v$-regular, i.e., all vertices have the same valence $v$.
Then if the $x_1,\ldots,x_n$ are interpreted as the roots of a polynomial of degree $n$,
or after homogenization as a homogeneous polynomial of degree $n$ in two variables,
i.e., a binary form, your sum defines an $SL_2$ invariant for such a binary form.
The nonregular case likewise corresponds to what 19th century mathematicians called
covariants.

Here is a fact. In the regular case, the product $nv$ has to be even since this is the
number of edges. Take $n=5$ and $v$ even but not divisible by 4 such that $v<18$. Then for every graph satisfying
this condition the polynomial is identically zero.
Likewise you can take $v=5$ and impose $n$ even but not divisible by 4 and $n<18$ and the result also is that
all graphs of this kind give zero. This is a nontrivial fact which has to do with
the invariants of the binary quintic: there is no skew invariant before degree 18 which is
the degree of Hermite's invariant.

Another example of your question is the following. Take $n=m^2$, and arrange the vertices
into an $m\times m$ square array. Take for the graph $g$ the following:
put an edge between two vertices if they are in the same row or in the same column.
It is trivial to see that the polynomial will vanish if $m$ is odd.
Now if you can prove that the graph polynomial is nonzero (for any even $m$)
then you would have proved the Alon-Tarsi conjecture, which implies the even case
of the Rota basis conjecture.
As far as I know these are widely open problems.